op amp circuit

Hello

I need a solution please. I have a Hall Effect sensor that I want to use in a magnetic lock to sense when the armature( strike plate) has mated with the stator. There is a change in the magnetic field when the armature couples with the stator. I want to make use of that property to show that the lock is secure.

The Hall Effect sensor, A1302, from Allegro is what I have in mind. It is a linear ratiometric device. The quiescent output sits at half supply voltage, in this case 2.5V with a 5V supply. The output varies up and down as the magnetic fields change. This change is very small, typically 1.35mV/G. I can see the variations on my scope but I need to make the changes bigger. This is where I need an op-amp expert.

I want to make the 2.5V offset move down to 0V and only amplify the positive movement so that I have a range that goes from 0V to about 5V for the small change from the sensor. I cannot use an AC amplifier as this is a changing DC voltage only.

I am stumped by this and would appreciate some help if you are prepared to do that.

Thank you. Adrian Carboni Adco Electronics South Africa

Any regular op amp (that will run from 5V supplies) will do the job. The LM358 would be about the cheapest.

There are better single supply op amps, but they may not help much.

The main problem is that your zero magnetic field output voltage - nominally half the supply voltage or 2.5V for a 5V supply - isn't all that well defined.

According the the Allegro A1301 data sheet, there's a +/-100mV tolerance on the zero-field output voltage.

Since the A1301 has a sensitivity of 2.5 +/-0.7 mV/G that's 40G of uncertainty (worst case 55G)

If the change in the magnetic field is bigger than that, you could get by with a pretty simple circuit.

In reality what you probably want to do it to use a comparator. The LM393 dual is about as cheap as they get, and works fine from 5V supplies.

Use a 500R potentiometer - 19mm 20 or 25-turn parts are stable and easy to set up - with 3k resistors from either end to the 5V and 0V rails respectively - to set up your nominally 2.5V reference voltage

Hook up the output from your A1301 to the non-inverting input of the on of the two comparators in the LM393 package.

Hook up the comparator output to the 5V through a 4k7 pull-up resistor (the LM393 has an open collector output, and won't work without some kind of pull-up resistor) and also connect the output to the wiper of the potentiometer (and the non-inverting input of the comparator) through a 1.5M resistor which will give about 5mV of hysterisis at switching.

Twiddle the potentiometer to a voltage about halfway between the A1301's output with the lock open and its output with the lock closed, and you should be okay, unless that difference is less than the 5mV hysterisis of the circuit, in which case you could certainly forget the whole idea.

In practice, you would have to measure a lot of "lock open" versus "lock closed" voltages, for different locks, for different A1301's and at different temperatures before you could be certain that the approach could work at all, let alone work with the minimal circuit I've (vaguely) described.

E-mail me with more details (at snipped-for-privacy@ieee.org) if you want a more specific circuit description - LTSpice circuit diagrams are easy to e-mail.

-- Bill Sloman, Nijmegen

If you've got + and - supply rails, then an instrument amp (AD620, INA121, or maybe some cheaper offering) with a pot to give you 2.5 volts on one input will work. Adjust gain to taste.

(You can do this lots of other ways... inverting opamp with -2.5 volt signal (from pot) into summing junction for instance.)

George H.

In the 1970-72 timeframe, I was analyzing relays to determine _energy_ required to close them, and energy to be absorbed (from the coil) when they opened. I don't remember much about it now (haven't used a relay since :-), but I do remember that the _inductance_ of the coil is dramatically higher when the armature is "pulled-in". ...Jim Thompson

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| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
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I love to cook with wine.     Sometimes I even put it in the food.

The GMR sensors from

have a much higher output than Hall effect sensors and are lower power. Maybe worth a look.

Cheers

--
Syd

Further musing. The mechanics of how you mount the Hall device relative to the pole-piece and armature can make all the difference in the world with regard to signal amplitude.

Draw us a picture. ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.

What is the problem that you could not use a high impedance voltmeter (10Me= g) to tell us the change in output voltage from the sensor, as well as the = VCC supply voltage, all to 3 digits minimum, in the two different lock posi= tions?

If your lock is exposed to any kind of temperature extremes , such as outdo= ors, then you can expect the nominal 2.5V output bias to vary over 2.2 to 2= .8 V range, and this is assuming Vcc stays locked onto 5.0V. So you can see= how a high gain configuration will have no way of knowing if it's amplifyi= ng a legitimate output change due to magnetic field or some variation due t= o environmental parameters.

to tell us the change in output voltage from the sensor, as well as the VCC supply voltage, all to 3 digits minimum, in the two different lock positions?

then you can expect the nominal 2.5V output bias to vary over 2.2 to 2.8 V range, and this is assuming Vcc stays locked onto 5.0V. So you can see how a high gain configuration will have no way of knowing if it's amplifying a legitimate output change due to magnetic field or some variation due to environmental parameters.

Yep. That's why I asked physical location. The best way to improve noise margin is to increase the signal _at_the_source_. ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.

It's all coming back to me... I was needing also to know how much _force_ the armature could pull. The force versus armature position can be derived from the inductance versus position... I'm vaguely remembering math functions like Hamiltonian and Lagrangian.

I was always amazed at the stuff I learned in the mechanical engineering classes. (I, as a member of MIT Course VI-B, honors, was required to take electives in other engineering and physics departments as if I was a major in that field :-( ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.

On the assumption of zero losses, the rate of work done against the restoring force of the spring and friction, plus the rate of work done on the magnetic field (LI delta I) equals the electrical power supplied. If you know the coil resistance and the spring constant, you can put in the electrical losses by hand, and figure out the rest.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510
845-480-2058

hobbs at electrooptical dot net
http://electrooptical.net

Yep. It's all coming back to me, though it's been 40 years ;-) ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.

Reminds me of the term "variable reluctance".

put 2.5V on the inverting input of the op-amp through a 5K resistor (eg 10k to +5v 10k to 0v) connect the non-inverting input to the sensor output put a resistor of the 5K*gain_desired between the output and inverting input

But first I'd be looking at sensor placement. with such a large electromagnet that sensor should give plenty of output in the right location.

--
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output sits at half

So, if you want to compare the key-inserted output with some known value, use a comparator with an appropriate threshold set. Or, two comparators will let you determine a range. There's no need for any intermediate step of amplification, that I can see. One does need to convert an analog value to a true/false value.

A comparator is a simplified operational amplifier, with logic-level output, and true differential input, and high gain, but without stability in a negative feedback circuit.

Alternately (if you are comfortable with microprocessors) you could send a ramp voltage to a bunch of comparators' (-) terminals, and determine by the time-of-switching on the comparator outputs what the voltages are on all the (+) terminals. You could also directly read the voltages on the Hall sensors with an A/D converter, lots of micros have multiple A/D channels built in.